1. ENERGY CONVERSION ONE (Course 25741) CHAPTER SEVEN INDUCTION MOTORS … (Maximum Torque…)

2. INDUCTION MOTOR MAXIMUM TORQUE maximum power transfer occurs when: R 2 /s=√R TH ^2 + (X TH +X 2 )^2 (1) solving (1) for slip  s max =R 2 / √R TH ^2 + (X TH +X 2 )^2 (2) Note: slip of rotor (at maximum torque) ~ R 2 rotor resistance applying this value of slip to torque equation 

3. INDUCTION MOTOR MAXIMUM TORQUE This maximum torque ~ V TH ^2 (or square of supply voltage) & inversely related to stator Impedances & rotor reactance The smaller a machine’s reactance the larger its maximum torque Note: s max ~ R 2, however maximum torque is independent of R 2 Torque-speed characteristic of a wound-rotor induction motor shown if figure next

4. INDUCTION MOTOR MAXIMUM TORQUE Effect of varying rotor resistance on T-ω of wound rotor

5. INDUCTION MOTOR MAXIMUM TORQUE as the value of external resistor connected to rotor circuit of a wound rotor through slip rings is increased the pullout speed decreased, however the maximum torque remains constant Advantage can be taken from this characteristic of wound-rotor induction motors to start very heavy loads If a resistance inserted into rotor circuit, T max can be adjusted to at starting conditions And while load is turning, extra resistance can be removed from circuit, & T max move up to near synchronous speed for regular operation

6. INDUCTION MOTOR EXAMPLE(1) A 2 pole, 50 Hz induction motor supplies 15kW to a load at a speed of 2950 r/min. What is the motor’s slip? What is the induced torque in the motor in Nm under these conditions? What will the operating speed of the motor be if its torque is doubled? How much power will be supplied by the motor when the torque is doubled?

7. INDUCTION MOTOR EXAMPLE(1)-SOLUTION (a) n sync = 120f e /p= 120x50/2=3000 r/min s= 3000-2950/3000=0.0167 or 1.67% (b) T ind =P conv /ω m =15 / (2950)(2πx1/60)=48.6 N.m. (c) as shown, in low slip region, torque-speed is linear & induced torque ~ s  doubling T ind slip would be 3.33 %  n m =(1-s)n sync =(1-0.0333)(3000)=2900 r/min (d) P conv =T ind ω m =97.2 x 2900 x 2πx1/60=29.5 kW

8. INDUCTION MOTOR EXAMPLE(2) A 460V, 25hp, 60Hz, 4-pole, Y-connected wound rotor induction motor has the following impedances in ohms per-phase referred to the stator circuit: R1 = 0.641 ΩR2 = 0.332 Ω X1 = 1.106 ΩX2 = 0.464 ΩXm = 26.3 Ω –What is the max torque of this motor? At what speed and slip does it occur? –What is the starting torque? –When the rotor resistance is doubled, what is the speed at which the max torque now occurs? What is the new starting torque?

9. INDUCTION MOTOR EXAMPLE(2)-SOLUTION Thevenin Voltage : = = 266/ √(0.641)^2+(1.106+26.3)^2= 255.2 V =(0.641)(26.3/[1.106+26.3])^2=0.59 Ω X TH ≈X 1 =1.106 Ω (a)s max = R 2 / √R TH ^2 + (X TH +X 2 )^2 =0.332/√(0.59)^2+(1.106+0.464)^2=0.198

10. INDUCTION MOTOR EXAMPLE(2)-SOLUTION This corresponds to a mechanical speed of : n m =(1-s)n sync =(1-0.198)(1800)=1444 r/min the torque at this speed : = 3(255.2)^2 / {2x188.5x[0.59+√0.59^2+(1.106+0.464)^2]} =229 N.m.

11. INDUCTION MOTOR EXAMPLE(2)-SOLUTION (b) starting torque of motor found by s=1 = 3x255.2^2 x 0.332 / {188.5x[(0.59+0.332)^2+(1.106+0.464)^2]}=104 N.m. (c) rotor resistance is doubled,  s at T max doubles s max =0.396, and the speed at T max is: n m =(1-s)n sync =(1-0.396)(1800)=1087 r/min Maximum torque is still: T max =229 N.m. and starting torque is : T start =3(255.2)(0.664) / {(188.5)[(0.59+0.664)^2+(1.106+0.464)^2]} =170 N.m.

12. INDUCTION MOTOR VARIATION IN TORQUE-SPEED Discussion:

13. INDUCTION MOTOR VARIATION IN TORQUE-SPEED Desired Motor Characteristic Should behave: like the high-resistance wound- rotor curve; at high slips, & like the low- resistance wound-rotor curve at low slips

14. INDUCTION MOTOR VARIATION IN TORQUE-SPEED Control of Motor Characteristics by Cage Rotor Design: Leakage reactance X 2 represents the referred form of the rotor’s leakage reactance (reactance due to the rotor’s flux lines that do not couple with the stator windings.) Generally, the farther away the rotor bar is from the stator, the greater its X 2, since a smaller percentage of the bar’s flux will reach the stator. Thus, if the bars of a cage rotor are placed near the surface of the rotor, they will have small leakage flux and X 2 will be small.

15. INDUCTION MOTOR VARIATION IN TORQUE-SPEED Laminations from typical cage induction motor, cross section of the rotor bars: NEMA class A – large bars near the surface NEMA class B – large, deep rotor bars NEMA class C – double-cage rotor design NEMA class D – small bars near the surface

16. INDUCTION MOTOR TORQUE-SPEED of CLASS:A,B,C,D NEMA (National Electrical Manufacturers Association) class A Rotor bars are quite large and are placed near the surface of the rotor Low resistance (due to its large cross section) and a low leakage reactance X 2 (due to the bar’s location near the stator) Because of the low resistance, the pullout torque will be quite near synchronous speed ; full load slip less than 5% Motor will be quite efficient, since little air gap power is lost in the rotor resistance. ; However, since R 2 is small, starting torque will be small, and starting current will be high This design is the standard motor design Typical applications : driving fans, pumps, and other machine tools Principal problem: extremely high inrush current on starting, 500 to 800 % of rated

17. INDUCTION MOTOR TORQUE-SPEED of CLASS:A,B,C,D NEMA Class B At the upper part of a deep rotor bar, the current flowing is tightly coupled to the stator, and hence the leakage inductance is small in this region. Deeper in the bar, the leakage inductance is higher At low slips, the rotor’s frequency is very small, and the reactances of all the parallel paths are small compared to their resistances. The impedances of all parts of the bar are approx equal, so current flows through all the parts of the bar equally. The resulting large cross sectional area makes the rotor resistance quite small, resulting in good efficiency at low slips. At high slips (starting conditions), the reactances are large compared to the resistances in the rotor bars, so all the current is forced to flow in the low-reactance part of the bar near the stator. Since the effective cross section is lower, the rotor resistance is higher. Thus, the starting torque is relatively higher and the starting current is relatively lower than in a class A design (about 25% less) Applications similar to class A, and this type B have largely replaced type A Pullout Torque greater than or equal 200% of rated load torque

18. INDUCTION MOTOR TORQUE-SPEED of CLASS:A,B,C,D NEMA Class C It consists of a large, low resistance set of bars buried deeply in the rotor and a small, high-resistance set of bars set at the rotor surface. It is similar to the deep- bar rotor, except that the difference between low-slip and high-slip operation is even more exaggerated At starting conditions, only the small bars are effective, and the rotor resistance is high. Hence, high starting torque. However, at normal operating speeds, both bars are effective, and the resistance is almost as low as in a deep-bar rotor Used in high starting torque loads such as loaded pumps, compressors, and conveyors

19. INDUCTION MOTOR TORQUE-SPEED of CLASS:A,B,C,D NEMA class D Rotor with small bars placed near the surface of the rotor (higher-resistance material) High resistance (due to its small cross section) and a low leakage reactance X 2 (due to the bar’s location near the stator) Like a wound-rotor induction motor with extra resistance inserted into the rotor Because of the large resistance, the pullout torque occurs at high slip, and starting torque will be quite high, and low starting current (starting T, 275% T rated ) Typical applications : extremely high-inertia type loads

20. INDUCTION MOTOR TORQUE-SPEED of CLASS:A,B,C,D NEMA Class E and F Class E and Class F are already discontinued They are low starting torque machines These called soft-start induction motors These are also distinguished by having very low starting currents & used for starting-torque loads in situations where starting current were a problem

21. INDUCTION MOTOR TORQUE-SPEED of CLASS:A,B,C,D T-speed Curve for Different Rotor Design

22. INDUCTION MOTOR TORQUE-SPEED of CLASS:A,B,C,D Basic concepts of developing variable rotor resistance by deep rotor bars or double- cage rotors

23. INDUCTION MOTOR TORQUE-SPEED of CLASS:A,B,C,D Basic Concept continued; (Last Figure) In Fig (a): for a current flowing in the upper part of the deep rotor bar, the flux is tightly linked to the stator, and leakage L is small In Fig (b): current flowing at the bottom part of the bar, the flux is weakly linked to the stator, and leakage L is large Fig (c): Resulting equivalent circuit Since all parts of rotor bar are in parallel electrically, bar represents a series of parallel electric circuits, upper ones have smaller inductance & lower ones larger inductance : L<L 1 <L 2 <L 3

24. INDUCTION MOTOR EXAMPLE-3 A 460 V, 30 hp, 60 Hz, 4 pole, Y connected induction motor has two possible rotor designs: - A single cage rotor & - A double-cage rotor (stator identical for both designs) Single-cage modeled by: R 1 =0.641Ω, R 2 =0.3Ω X 1 =0.75 Ω, X 2 =0.5 Ω, X M =26.3 Ω Double-cage; modeled as tightly coupled high resistance outer cage in parallel with a loosely coupled, low-resistance inner cage, stator magnetization resistance & reactances identical R 2o =3.2 Ω X 2o =0.5 Ω (of outer-cage) R 2i =0.4 Ω X 2i =3.3 Ω (of outer-cage) Calculate torque-speed characteristics associated with two rotor designs solution by MATLAB Results: double-cage: slightly higher slip, smaller T max, higher T starting,

25. TRENDS IN INDUCTION MOTOR DESIGN Smaller motor for a given power output, great saving (modern 100 hp same size of 7.5 hp motor of 1897) However not necessarily increase in efficiency (used since electricity was inexpensive) New lines of high efficiency induction motors being produced by all major manufacturers using some the following techniques; 1- More copper in stator windings; reduce copper losses 2- rotor & stator length increased to reduce B in air gap (decreasing saturation and core loss) 3-More steel in stator, greater amount of heat transfer 4- using special high grade steel with low hysteresis loss in stator 5- steel made of especially thin guage & high resistivity toreduce eddy current loss 6-rotor carefully machined to produce uniform air gap, reducing stray load losses

26. INDUCTION MOTORS STARTING An induction motor has the ability to start directly, however direct starting of an induction motor is not advised due to high starting currents, may cause dip in power system voltage; that across-the-line starting not acceptable for wound rotor, by inserting extra resistance can be reduced; this increase starting torque, but also reduces starting current For cage type, starting current vary widely depending primarily on motor’s rated power & on effective rotor resistance at starting conditions

27. INDUCTION MOTORS STARTING To determine starting current, need to calculate the starting power required by the induction motor. A Code Letter designated to each induction motor, which can be seen in figure 7-34, may represent this. (The starting code may be obtained from the motor nameplate) In example: for code letter A; factor of kVA/hp is between 0-3.15 (not include lower bound of next higher class)

28. INDUCTION MOTORS STARTING EXAMPLE: what is starting current of a 15 hp, 208 V, code letter F, 3 phase induction motor? Maximum kVA / hp is 5.6  max. starting kVA of this motor is S start =15 x 5.6 = 84 kVA the starting current is thus: I L =S start / [√3 V T ] = 84 / [√3 x 208] = 233 A Starting current may be reduced by a starting circuit: a- inductor banks b- resistor banks c-reduce motor’s terminal voltage by autotransformer

29. INDUCTION MOTORS STARTING Autotransformer starter: During starting 1 & 3 closed, when motor is nearly up to speed; those contacts opened & 2 closed Note: as starting current reduced proportional to decrease in voltage, starting torque decreased as square of applied voltage, therefore just a certain reduction possible if motor is to start with a shaft load attached

30. INDUCTION MOTORS STARTING A typical full-voltage (across-the-line) motor magnetic starter circuit

31. INDUCTION MOTORS STARTING Start button pressed, rely coil M energized, & N.O. contacts M 1,M 2,M 3 close Therefore power supplied to motor & motor starts Contacts M 4 also close which short out starting switch, allowing operator to release it (start button) without removing power from M relay When stop button pressed, M relay de- energized, & M contacts open, stopping motor

32. INDUCTION MOTORS STARTING A magnetic motor starter circuit has several built-in protective features: 1- short-circuit protection 2- overload protection 3- under-voltage protection Short-circuit protection provided by fuses F 1,F 2,F 3 If sudden sh. cct. Develops within motor causes a current (many times greater than rated current) flow; these fuses blow disconnecting motor from supply (however, sh. cct. by a high resistance or excessive motor loads will not be cleared by fuses)

33. INDUCTION MOTORS STARTING Overload protection for motor is provided “OL” relays which consists of 2 parts: an over load heater, and overload contacts when an induction motor overloaded, it is eventually damaged by excessive heating caused by high currents However this damage takes time & motor will not be hurt by brief periods of high current (such as starting current) Undervoltage protection is also provided by controller If voltage applied to motor falls too much, voltage applied to M relay also fall, & relay will de-energize The M contacts open, removing power from motor terminals

34. INDUCTION MOTORS STARTING 3 step resistive starter Similar to previous, except that there are additional components present to control Removal of starting resistors Relays 1TD, 2TD, & 3 TD are time-delay relay

35. INDUCTION MOTORS STARTING Start button is pushed in this circuit, M relay energizes and power is applied to motor as before Since 1TD, 2TD, & 3TD contacts are all open the full starting resistor in series with motor, reducing the starting current When M contacts close, notice that 1 TD relay is energized, however there is a finite delay before 1TD contacts close, cutting out part of starting resistance & simultaneously energizing 2TD relay After another delay, 2TD contacts close, cutting out second part of resistor & energizing 3TD relay Finally 3TD contacts close, & entire starting resistor is out of circuit

36. INDUCTION MOTOR SPEED CONTROL Induction motors are not good machines for applications requiring considerable speed control. The normal operating range of a typical induction motor is confined to less than 5% slip, and the speed variation is more or less proportional to the load Since P RCL = s P AG, if slip is made higher, rotor copper losses will be high as well There are basically 2 general methods to control induction motor’s speed: - Varying synchronous speed - Varying slip

37. INDUCTION MOTOR SPEED CONTROL n sync = 120 f e / p so the only ways to change n sync is (1) changing electrical frequency (2) changing number of poles slip control can be accomplished, either by varying rotor resistance, or terminal voltage of motor Speed Control by Pole Changing Two major approaches: 1- method of consequent poles 2- multiple stator windings

38. INDUCTION MOTOR SPEED CONTROL 1- method of consequent poles relies on the fact that number of poles in stator windings can easily changed by a factor of 2:1, with simple changes in coil connections - a 2-pole stator winding for pole changing. Very small rotor pitch  In next figure for windings of phase “a” of a 2 pole stator, method is illustrated

39. INDUCTION MOTOR SPEED CONTROL A view of one phase of a pole changing winding In fig(a), current flow in phase a, causes magnetic field leave stator in upper phase group (N) & enters stator in lower phase group (S), producing 2 stator magnetic poles

40. INDUCTION MOTOR SPEED CONTROL Now, if direction of current flow in lower phase group reversed, magnetic field leave stator in both upper phase group, & lower phase group, each will be a North pole while flux in machine must return to stator between two phase groups, producing a pair of consequent south magnetic poles (twice as many as before) Rotor in such a motor is of cage design, and a cage rotor always has as many poles as there are in stator when motor reconnected from 2 pole to 4 pole, resulting maximum torque is the same (for :constant-torque connection) half its previous value (for: square-law-torque connection used for fans, etc.), depending on how the stator windings are rearranged Next figure, shows possible stator connections & their effect on torque-speed

41. INDUCTION MOTOR SPEED CONTROL Possible connections of stator coils in a pole-changing motor, together with resulting torque-speed characteristics: (a) constant-torque connection : power capabilities remain constant in both high & low speed connections (b) constant hp connection: power capabilities of motor remain approximately constant in both high- speed & low-speed connections (c) Fan torque connection: torque capabilities of motor change with speed in same manner as fan-type loads Shown in next figure 

42. INDUCTION MOTOR SPEED CONTROL Figure of possible connections of stator coils in a pole changing motor (a)constant-torque Connection: torque capabilities of motor remain approximately constant in both high-speed & low-speed connection (b)Constant-hp connection: power capabilities of motor remain approximately constant in … (c)Fan torque connection:

43. INDUCTION MOTOR SPEED CONTROL Major Disadvantage of consequent-pole method of changing speed: speeds must be in ratio of 2:1 traditional method to overcome the limitation: employ multiple stator windings with different numbers of poles & to energize only set at a time Example: a motor may wound with 4 pole & a set of 6 pole stator windings, then its sync. Speed on a 60 Hz system could be switched from 1800 to 1200 r/min simply by supplying power to other set of windings however multiple stator windings increase expense of motor & used only it is absolutely necessary Combining method of consequent poles with multiple stator windings a 4 –speed motor can be developed Example: with separate 4 & 6 pole windings, it is possible to produce a 60 Hz motor capable of running at 600, 900, 1200, and 1800 r/min

44. INDUCTION MOTOR SPEED CONTROL Speed Control by Changing Line Frequency Changing the electrical frequency will change the synchronous speed of the machine Changing the electrical frequency would also require an adjustment to the terminal voltage in order to maintain the same amount of flux level in the machine core. If not the machine will experience (a) Core saturation (non linearity effects) (b) Excessive magnetization current

45. INDUCTION MOTOR SPEED CONTROL Varying frequency with or without adjustment to the terminal voltage may give 2 different effects : (a) Vary frequency, stator voltage adjusted – generally vary speed and maintain operating torque (b) Vary Frequency, stator voltage maintained – able to achieve higher speeds but a reduction of torque as speed is increased There may also be instances where both characteristics are needed in the motor operation; hence it may be combined to give both effects With the arrival of solid-state devices/power electronics, line frequency change is easy to achieved and it is more flexible for a variety of machines and application Can be employed for control of speed over a range from a little as 5% of base speed up to about twice base speed

46. INDUCTION MOTOR SPEED CONTROL Running below base speed, the terminal voltage should be reduced linearly with decreasing stator frequency This process called derating, failing to do that cause saturation and excessive magnetization current (if f e decreased by 10% & voltage remain constant  flux increase by 10% and cause increase in magnetization current) When voltage applied varied linearly with frequency below base speed, flux remain approximately constant, & maximum torque remain fairly high, therefore maximum power rating of motor must be decreased linearly with frequency to protect stator cct. From overheating Power supplied to : √3 V L I L cosθ should be decreased if terminal voltage decreased Figures (7-42 )

47. INDUCTION MOTOR SPEED CONTROL Variable-frequency speed control (a)family of torque-speed characteristic curves for speed below base speed (assuming line voltage derated linearly with frequency (b) Family of torque-speed characteristic curves for speeds above base speed, assuming line voltage held constant

48. INDUCTION MOTOR SPEED CONTROL Speed control by changing Line Voltage Torque developed by induction motor is proportional to square of applied voltage Varying the terminal voltage will vary the operating speed but with also a variation of operating torque In terms of the range of speed variations, it is not significant hence this method is only suitable for small motors only

49. INDUCTION MOTOR SPEED CONTROL Variable-line-voltage speed control

50. INDUCTION MOTOR SPEED CONTROL Speed control by changing rotor resistance In wound rotor, it is possible to change the torque-speed curve by inserting extra resistances into rotor cct. However, inserting extra resistances into rotor cct. seriously reduces efficiency Such a method of speed control normally used for short periods, to avoid low efficiency

51. INDUCTION MOTOR CCT MODEL PAR. MEAURSEMENT Determining Circuit Model Parameters R 1,R 2,X 1,X 2 and X M should be determined Tests (O.C. & S.C.) performed under precisely controlled conditions Since resistances vary with Temperature & rotor resistance also vary with rotor frequency Exact details described in IEEE standards 112 Although details of tests very complicated, concepts behind them straightforward & will be explained here

52. INDUCTION MOTOR CCT MODEL PAR. MEAURSEMENT No Load Test Measures rotational losses & provides information about magnetization current Test cct. shown in (a), motor allowed to spin freely Wattmeters, a voltmeter and 3 ammeters

53. INDUCTION MOTOR CCT MODEL PAR. MEAURSEMENT In this test, only load mechanical losses, & slip very small (as 0.001 or less) Equivalent cct. shown in figure (b) Resistance corresponding to power conversion is R 2 (1-s)/s much larger than R 2 & much larger than X 2 so eq. cct. Reduces to last in (b) output resistor in parallel with magnetization reactance X M & core losses R C Input power measured by meters equal losses, while rotor copper losses negligible (I 2 extremely small), P SL =3I 1 ^2 R 1

54. INDUCTION MOTOR CCT MODEL PAR. MEAURSEMENT P in =P SCL +P core +P F&W +P misc =3 I 1 ^2 R 1 + P rot So eq. cct. In this condition contains R C and R 2 (1-s)/s in parallel with X M While current to provide magnetic field is large due to high reluctance of air gap & so X M would be much smaller than resistance in parallel with it Overall P.F. very small with large lagging current : |Z eq |=V φ /I 1,nl ≈ X 1 +X M if X 1 known by another fashion, X M can be determined

55. INDUCTION MOTOR DC Test for STATOR RESISTANCE The locked-rotor test later used to determine total motor circuit resistance However to determine rotor resistance R 2 that is very important and affect the torque-speed curve, R 1 should be known There is a dc test for determining R 1. a dc power supply is connected to two of 3 terminals of a Y connected induction motor Current adjusted to rated value & voltage between terminals measured

56. INDUCTION MOTOR DC Test for STATOR RESISTANCE reason for setting current to rated value is to heat windings to same temperature of normal operation 2R 1 = V DC /I DC or R 1 =V DC /[2 I DC ] With R 1, stator copper losses at no load can be determined rotational losses determined as difference of P in at no load & stator copper loss R 1 determined by this method is not accurate, due to neglect of skin effect using an ac voltage

57. INDUCTION MOTOR LOCKED ROTOR TEST Third test to determine cct. Parameters of an induction motor is called : locked-rotor test In this test rotor is locked & cannot move Voltage applied to motor, voltage, current & power are measured An ac voltage applied to stator, current flow adjusted to full-load value Then, voltage, current, & power flowing to motor measured

58. INDUCTION MOTOR LOCKED ROTOR TEST Since rotor is stationary, slip s=1. & R 2 /s equal R 2 (small value) Since R 2 & X 2 so small, all input current will flow through them rather X M and circuit is a series of X 1,R 1,X 2 and R 2 There is one problem with this test  in normal operation, stator frequency is line frequency (50 or 60 Hz) At starting conditions, rotor also at power frequency (while in normal operation slip 2 to 4 % and frequency 1 to 3 Hz) & it does not simulate normal operation A compromise : is to use a frequency 25% or less of rated frequency

59. INDUCTION MOTOR LOCKED ROTOR TEST This acceptable for constant resistance rotors (design class A and D) it leaves a lot to be desired when looking for normal rotor resistance of a variable resistance rotor a great deal of care required taking measurement for these tests a test voltage & frequency set up, current flow in motor quickly adjusted to about rated voltage, & input power, voltage and current measured before motor heat up

60. INDUCTION MOTOR LOCKED ROTOR TEST P= √3 V T I L cos θ So locked-rotor P.F. found as: PF = cosθ= P in / [√3 V T I L ] Impedance angle is θ=acos P.F. Magnitude of total impedance : |Z LR | =Vφ / I 1 = V T /[√3 I L ] Angle of total impedance is θ, therefore, Z LR =R LR +jX’ LR = |Z LR | cos θ +j |Z LR | sinθ Locked-rotor resistance R LR =R 1 +R 2 While locked-rotor reactance X’ LR =X’ 1 +X’ 2 Where X’ 1 and X’ 2 are stator & rotor reactances at test frequency

61. INDUCTION MOTOR LOCKED ROTOR TEST Rotor resistance R 2 can be found: R 2 =R LR -R 1 R 1 determined in dc test Total rotor reactance referred to stator can be found Since reactance ~ f  total eq. reactance at normal operating frequency: X LR =f rated / f test = X 1 +X 2 No simple way for separation of stator & rotor contributions Experience, shown motors of certain design have certain proportions between rotor & stator reactances

62. INDUCTION MOTOR LOCKED ROTOR TEST Table summarizes this experience In normal practice does not matter how X LR is divided, since reactance appears as X 1 +X 2 in all torque equations X 1 &X 2 as function of X LR Rotor Design X 1 X 2 Wound rotor 0.5 X LR Design A 0.5 X LR Design B 0.4 X LR 0.6 X LR Design C 0.3 X LR 0.7 X LR Design D 0.5 X LR

63. INDUCTION MOTOR Equivalent CCT Parameters-Example Following test data taken on a 7.5 hp, 4 pole, 208 V, 60 Hz, design A, Y connected induction motor with a rated current of 28 A. Dc test result: V DC =13.6 I DC = 28.0 A No load test: V T =208 V f =60 Hz I A =8.12 A, I B =8.2 A, I C =8.18 A P in =420 W Locked rotor test: V T =25 V f=15 Hz I A =28.1 A, I B =28.0 A, I C =27.6 A P in =920 W

64. INDUCTION MOTOR Equivalent CCT Parameters-Example (a)sketch per phase equivalent circuit of this motor (b) Find slip at pullout torque, and find the value of pullout torque Solution : (a)from dc test  R 1 =V DC /[2I DC ] =13.6/[2x28]= 0.243 Ω from no load test: I L,av =[8.12+8.2+8.18]/3=8.17 A V φ,nl =208/√3 = 120 V therefore: |z nl |=120/8.17=14.7 Ω =X 1 +X M when X 1 is known, X M can be found

65. INDUCTION MOTOR Equivalent CCT Parameters-Example The copper losses: P SCL =3I 1 ^2 R 1 =3 X 8.17^2 x 0.243 Ω =48.7 W No load rotational losses are: P rot =P in,nl – P SCL,nl =420 -48.7 =371.3 W from locked-rotor test: I L,av =[28.1+28.0+27.6]/3=27.9 A Locked rotor impedance is: |Z LR |=25/[√3x27.9]=0.517 Ω Impedance angle θ=acos 920/[√3 x 25 x 27.9]=acos 0.762=40.4 ◦  R LR =0.517cos 40.4=0.394 Ω =R 1 +R 2 since R 1 =0.243 Ω  R 2 =0.151 Ω The reactance at 15 Hz ; X’ LR =0.517 sin 40.4=0.335Ω

66. INDUCTION MOTOR Equivalent CCT Parameters-Example Equivalent reactance at 60 Hz: X LR = f rated /f test X’ LR =60/15 x 0.335 =1.34 Ω For design class A, this reactance divided equally between rotor & stator: X 1 =X 2 =0.67 Ω X M =|Z nl |-X 1 =14.7-0.67=14.03 Ω Per phase equivalent cct shown below:

67. INDUCTION MOTOR Equivalent CCT Parameters-Example (b) for this equivalent cct. Thevenin equivalent found as follows: V TH =114.6 V, R TH =0.221 Ω, X TH =0.67 Ω  slip at pullout torque is : s max = R 2 /[√ R TH ^2+(X TH +X 2 )^2]= 0.151/√0.243^2+(0.67+0.67)^2=0.111=11.1% Maximum torque of this motor is given by: T max = 3 V TH ^2/{2ω sync [R TH +√R TH ^2+(X TH +X 2 )^2]}= 3 x 114.6 ^2 /{2x188.5x[0.221+√0.221^2+(0.2x0.67)^2]}=66.2 N.m.

68. INDUCTION GENERATOR The torque –speed curve shown when induction motor driven at speed greater than n sync by a prime mover, direction of induced torque reverses & act as a generator